Some Basic Concepts of Chemistry - Result Question 51
50. Calculate the molality of $1.0 L$ solution of $93 % H _2 SO _4$, (weight/volume). The density of the solution is $1.84 g / mL$.
(1990, 1M)
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Answer:
Correct Answer: 50. (10.42)
Solution:
- $93 % H _2 SO _4$ solution weight by volume indicates that there is $93 g H _2 SO _4$ in $100 mL$ of solution.
If we consider $100 mL$ solution, weight of solution $=184 g$
Weight of $H _2 O$ in $100 mL$ solution $=184-93=91 g$
$\Rightarrow$ Molality $=\frac{\text { Moles of solute }}{\text { Weight of solvent }(g)} \times 1000$
$=\frac{93}{98} \times \frac{1000}{91}=10.42$
BETA
